Simplify the following expression: $y = \dfrac{-7x^2- 27x+40}{-7x + 8}$
First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(-7)}{(40)} &=& -280 \\ {a} + {b} &=& &=& {-27} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $-280$ and add them together. Remember, since $-280$ is negative, one of the factors must be negative. The factors that add up to ${-27}$ will be your ${a}$ and ${b}$ When ${a}$ is ${8}$ and ${b}$ is ${-35}$ $ \begin{eqnarray} {ab} &=& ({8})({-35}) &=& -280 \\ {a} + {b} &=& {8} + {-35} &=& -27 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({-7}x^2 +{8}x) + ({-35}x +{40}) $ Factor out the common factors: $ x(-7x + 8) + 5(-7x + 8)$ Now factor out $(-7x + 8)$ $ (-7x + 8)(x + 5)$ The original expression can therefore be written: $ \dfrac{(-7x + 8)(x + 5)}{-7x + 8}$ We are dividing by $-7x + 8$ , so $-7x + 8 \neq 0$ Therefore, $x \neq \frac{8}{7}$ This leaves us with $x + 5; x \neq \frac{8}{7}$.